Derivatives of Inverse Trigonometric Functions

Basic Inverse Trig Derivatives

With Chain Rule

For $u = u(x)$:

Key Relationships

Note that:

• $\frac{d}{dx}[\arcsin(x)] = -\frac{d}{dx}[\arccos(x)]$
• $\frac{d}{dx}[\arctan(x)] = -\frac{d}{dx}[\text{arccot}(x)]$
• $\frac{d}{dx}[\text{arcsec}(x)] = -\frac{d}{dx}[\text{arccsc}(x)]$

This is because $\arcsin(x) + \arccos(x) = \frac{\pi}{2}$ (and similarly for the other pairs).

Derivation of arctan(x)

If $y = \arctan(x)$, then $x = \tan(y)$.

Differentiating implicitly:

$1 = \sec^2(y) \cdot \frac{dy}{dx}$

$\frac{dy}{dx} = \frac{1}{\sec^2(y)} = \cos^2(y)$

Since $\tan(y) = x$ and $\sec^2(y) = 1 + \tan^2(y)$:

$\frac{dy}{dx} = \frac{1}{1 + \tan^2(y)} = \frac{1}{1 + x^2}$

Examples

Example 1: Basic arcsin

$f(x) = \arcsin(2x)$

$f'(x) = \frac{2}{\sqrt{1 - (2x)^2}} = \frac{2}{\sqrt{1 - 4x^2}}$

Example 2: arctan with chain rule

$f(x) = \arctan(x^2)$

$f'(x) = \frac{2x}{1 + (x^2)^2} = \frac{2x}{1 + x^4}$

Example 3: Product with inverse trig

$f(x) = x \cdot \arcsin(x)$

$f'(x) = x \cdot \frac{1}{\sqrt{1 - x^2}} + \arcsin(x) \cdot 1 = \frac{x}{\sqrt{1 - x^2}} + \arcsin(x)$

Example 4: Composition

$f(x) = \arctan(e^x)$

$f'(x) = \frac{e^x}{1 + e^{2x}}$

Example 5: arccos

$f(x) = \arccos(1 - x^2)$

$f'(x) = -\frac{-2x}{\sqrt{1 - (1 - x^2)^2}} = \frac{2x}{\sqrt{1 - 1 + 2x^2 - x^4}}$

$= \frac{2x}{\sqrt{2x^2 - x^4}} = \frac{2x}{|x|\sqrt{2 - x^2}}$

Example 6: Sum of inverse functions

$f(x) = \arcsin(x) + \arccos(x)$

$f'(x) = \frac{1}{\sqrt{1 - x^2}} + \left(-\frac{1}{\sqrt{1 - x^2}}\right) = 0$

This confirms that $\arcsin(x) + \arccos(x) = \text{constant} = \frac{\pi}{2}$.